How To Solve A System With 3 Variables

How to Solve a System of Equations with Three Variables

Solving a system of equations with three variables might seem daunting, but with a systematic approach and a solid understanding of the underlying principles, it becomes a manageable task. This comprehensive guide will walk you through various methods, providing you with the tools and techniques to tackle these problems confidently. We'll cover substitution, elimination, and using matrices – offering a diverse toolkit to suit different equation types and personal preferences.

Understanding Systems of Three Variables

Before diving into the solutions, let's establish a firm grasp of what we're dealing with. A system of three variables involves three linear equations, each containing three unknowns (typically represented as x, y, and z). The goal is to find the values of x, y, and z that satisfy all three equations simultaneously. This point (x, y, z) represents the intersection of the three planes defined by the equations in three-dimensional space.

There are three possible outcomes when solving a system of three linear equations:

• Unique Solution: The three planes intersect at a single point. This means there is one unique solution (x, y, z) that satisfies all three equations. This is the most common scenario.

• No Solution: The planes are parallel or intersect in such a way that there is no point common to all three. In this case, the system is inconsistent, and there is no solution.

• Infinite Solutions: The planes intersect along a line or coincide completely. This implies that there are infinitely many solutions, satisfying a specific relationship between x, y, and z.

Method 1: Elimination Method

The elimination method, also known as the addition method, is a powerful technique for solving systems of equations. The core idea is to eliminate one variable at a time by adding or subtracting equations strategically. Let's illustrate with an example:

Example: Solve the following system of equations:

1. x + y + z = 6
2. 2x - y + z = 3
3. x + 2y - z = 3

Steps:

1. Choose a variable to eliminate: Let's eliminate 'z' first. Notice that equations 1 and 3 have opposite signs for 'z'. Adding these two equations directly eliminates 'z':

   (x + y + z) + (x + 2y - z) = 6 + 3 => 2x + 3y = 9 (Equation 4)

2. Eliminate the same variable again: We need another equation without 'z'. Let's use equations 1 and 2. Subtracting equation 1 from equation 2 eliminates 'z':

   (2x - y + z) - (x + y + z) = 3 - 6 => x - 2y = -3 (Equation 5)

3. Solve the resulting system: Now we have a system of two equations with two variables (x and y):

   2x + 3y = 9 x - 2y = -3

   We can solve this using either elimination or substitution. Let's use elimination. Multiply equation 5 by 2 to eliminate x:

   2(x - 2y) = 2(-3) => 2x - 4y = -6

   Subtract this new equation from equation 4:

   (2x + 3y) - (2x - 4y) = 9 - (-6) => 7y = 15 => y = 15/7

4. Substitute back: Substitute the value of y (15/7) into equation 5 to solve for x:

   x - 2(15/7) = -3 => x = -3 + 30/7 = 9/7

5. Solve for the remaining variable: Substitute the values of x (9/7) and y (15/7) into any of the original equations (let's use equation 1) to solve for z:

   (9/7) + (15/7) + z = 6 => 24/7 + z = 42/7 => z = 18/7

Solution: The solution to the system of equations is x = 9/7, y = 15/7, and z = 18/7.

Method 2: Substitution Method

The substitution method involves solving one equation for one variable in terms of the other two, and then substituting this expression into the other two equations. This reduces the system to two equations with two variables, which can then be solved using the same technique. This method is particularly useful when one equation can easily be solved for one variable.

Let's use the same example as above:

1. x + y + z = 6
2. 2x - y + z = 3
3. x + 2y - z = 3

Steps:

1. Solve for one variable: From equation 1, we can solve for x: x = 6 - y - z

2. Substitute: Substitute this expression for x into equations 2 and 3:

   2(6 - y - z) - y + z = 3 => 12 - 2y - 2z - y + z = 3 => -3y - z = -9 (Equation 4) (6 - y - z) + 2y - z = 3 => 6 + y - 2z = 3 => y - 2z = -3 (Equation 5)

3. Solve the system: Now solve the system of equations 4 and 5. We can solve equation 5 for y: y = 2z - 3. Substitute this into equation 4:

   -3(2z - 3) - z = -9 => -6z + 9 - z = -9 => -7z = -18 => z = 18/7

4. Back-substitute: Substitute z = 18/7 into y = 2z - 3 to find y: y = 2(18/7) - 3 = 15/7

5. Find the remaining variable: Substitute y = 15/7 and z = 18/7 into x = 6 - y - z to find x: x = 6 - (15/7) - (18/7) = 9/7

Solution: The solution is the same as before: x = 9/7, y = 15/7, z = 18/7.

Method 3: Using Matrices (Gaussian Elimination)

The matrix method, specifically Gaussian elimination, is a powerful and efficient technique for solving systems of linear equations, especially for larger systems. It involves representing the system of equations as an augmented matrix and then performing row operations to transform it into row-echelon form or reduced row-echelon form.

Let's use the same example:

1. x + y + z = 6
2. 2x - y + z = 3
3. x + 2y - z = 3

Steps:

1. Create the augmented matrix:

   [ 1  1  1 | 6 ]
   [ 2 -1  1 | 3 ]
   [ 1  2 -1 | 3 ]
   

2. Row operations: The goal is to transform the matrix into row-echelon form (upper triangular form) using row operations (swapping rows, multiplying a row by a constant, adding a multiple of one row to another).

   • R2 -> R2 - 2R1: Subtract twice the first row from the second row:

     [ 1  1  1 | 6 ]
     [ 0 -3 -1 |-9 ]
     [ 1  2 -1 | 3 ]
     

   • R3 -> R3 - R1: Subtract the first row from the third row:

     [ 1  1  1 | 6 ]
     [ 0 -3 -1 |-9 ]
     [ 0  1 -2 |-3 ]
     

   • R2 -> R2 / -3: Divide the second row by -3:

     [ 1  1  1 | 6 ]
     [ 0  1  1/3 | 3 ]
     [ 0  1 -2 |-3 ]
     

   • R3 -> R3 - R2: Subtract the second row from the third row:

     [ 1  1  1 | 6 ]
     [ 0  1  1/3 | 3 ]
     [ 0  0 -7/3 |-6 ]
     

   • R3 -> R3 * (-3/7): Multiply the third row by -3/7:

     [ 1  1  1 | 6 ]
     [ 0  1  1/3 | 3 ]
     [ 0  0  1 | 18/7 ]
     

3. Back-substitution: Now the matrix is in row-echelon form. We can use back-substitution to solve for the variables:

   • z = 18/7
   • y + (1/3)(18/7) = 3 => y = 15/7
   • x + (15/7) + (18/7) = 6 => x = 9/7

Solution: Again, the solution is x = 9/7, y = 15/7, z = 18/7.

Handling Inconsistent and Dependent Systems

Not all systems of three equations have a unique solution. If, during the elimination or Gaussian elimination process, you encounter a contradiction (e.g., 0 = 1), the system is inconsistent, and there is no solution. If you end up with a redundant equation (e.g., 0 = 0), the system is dependent, meaning there are infinitely many solutions. In the case of infinitely many solutions, you'll express the solution set in terms of a parameter (e.g., x = t, y = 2t + 1, z = 3t).

Practical Applications

Solving systems of three variables has widespread applications across various fields:

• Engineering: Analyzing circuit networks, structural mechanics, and fluid dynamics often involves solving systems of equations.

• Physics: Solving problems in classical mechanics, electromagnetism, and thermodynamics requires determining multiple unknown quantities simultaneously.

• Economics: Linear programming and input-output models use systems of equations to model economic relationships.

• Computer Graphics: 3D transformations and rendering rely on solving systems of equations to manipulate and display objects in virtual spaces.

• Chemistry: Balancing chemical equations and determining reaction stoichiometry involve solving systems of equations.

Conclusion

Solving a system of equations with three variables can be approached using various methods – elimination, substitution, or matrices. The choice of method often depends on the specific form of the equations and personal preference. By mastering these techniques and understanding the underlying principles, you'll be well-equipped to tackle these problems and apply them effectively in numerous real-world applications. Remember to always check your solution by substituting the values back into the original equations to ensure accuracy. Consistent practice is key to building proficiency in solving these systems.